Calculating proportional distribution among three individuals

Normalize the tracking parameters by converting everything to failure percentages . Total students = 100\%. Failed in English (E_f) = 100\% - 80\% = 20\%. Failed in Hindi (H_f) = 100\% - 70\% = 30\%. Failed in both subjects (E_f \cap H_f) = 15\% .

Use set theory to isolate unique failures: Failed in English only = 20\% - 15\% = 5\%. Failed in Hindi only = 30\% - 15\% = 15\%.

The question asks for students who failed in only one subject. Sum these distinct sets: Total = 5\% + 15\% = 20\%. Wait, let's re-verify the step or check if the prompt text specifies something else. Ah! Let's check total union: Total failed in at least one = 20 + 30 - 15 = 35\%. Let's re-read the exact question phrasing: "failed in only one subject". If English only is 5% and Hindi only is 15%, the sum is 20%. Let's review if the pass numbers imply something else when mapped as a pass union: Total passed at least one = 100\% - 15\% = 85\%. Using P(A \cup B) = P(A) + P(B) - P(A \cap B) \implies 85 = 80 + 70 - P(Pass Both) \implies P(Pass Both) = 65\%.

Sum of single-subject failures = 15\% + 5\% = 20\%.

Let's check why 35% is listed as an option. 35\% is the total union of failures (20 + 30 - 15 = 35\%). However, the specific prompt asks for those who failed in only one subject, which evaluates mathematically to 20%.

Answer: (b).